Question: Evaluate the improper integral if it exists. $\int^{\infty}_{0}\sin(x)\,dx $ Choose 1 answer: Choose 1 answer: (Choice A) A $-1$ (Choice B) B $0$ (Choice C) C $1$ (Choice D) D The improper integral diverges.
First, let's rewrite the improper integral: $\int^{\infty}_{0}\sin(x)\,dx=\lim_{b\to\infty}\int^{b}_{0}\sin(x)\,dx$ We can now evaluate the integral: $\begin{aligned} \phantom{\int_{0}^{\infty}\sin x\,dx}&=\lim_{b\to\infty}\int_{0}^{b}\sin( x)\,dx\\ \\ \\ &=\lim_{b\to\infty}\Big[-\cos x\Big]_{0}^b\\ \\ \\ &=\lim_{b\to\infty}(-\cos b+\cos 0)\\ \\ &=\lim_{b\to\infty}(-\cos b)+\lim_{b\to\infty}(\cos(0))\\\\ &=DNE+1\\ \\ &=DNE \end{aligned}$ The answer: The improper integral diverges.